Suppressing argparse Error message - Python

Question

I'm using argparse to parse command line arguments. However at this time, I have an unusual request: I want to suppress the error message. For example:

 # !/usr/bin/env python
 try:
        parser = argparse.ArgumentParser(description='Parse')
        parser.add_argument('url', metavar='URL', type=str, nargs='+', 
                            help='Specify URL')
 except:
        print("Invalid arguments or number of arguments")
        exit(2)

so I only expect that to print "Invalid arguments or number of arguments" and nothing else. However, then I execute the code for example:

./foo --BOGUS

I'm getting the full usage message:

usage: foo [-h]
foo: error: too few arguments
foo: Invalid arguments or number of arguments

How can I achieve that?

Eugene Yarmash · Accepted Answer · 2016-03-06 19:12:05Z

8

You could use contextlib.redirect_stderr to suppress the output from argparse:

import io
from contextlib import redirect_stderr

try:
    f = io.StringIO()
    with redirect_stderr(f):
        parser.parse_args()
except:
    print("Invalid arguments or number of arguments")
    exit(2)

Another option is to subclass ArgumentParser and override the exit method.

answered Mar 6, 2016 at 19:12

Eugene Yarmash

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Comments

RootTwo · Accepted Answer · 2016-03-06 22:11:15Z

4

ArgumentParser.parse_known_args(args=None, namespace=None) might help. It parses the command line and returns a tuple (namespace, unknown_args). namespace contains the args that were recognized and unknown_args is a list of args that weren't recognized.

ns, unknown_args = parser.parse_known_args()
if unknown_args != []:
    # do something with erroneous args, e.g., silent exit

else:
    # do something else

answered Mar 6, 2016 at 22:11

RootTwo

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Comments

Thomas · Accepted Answer · 2021-04-13 10:35:51Z

3

As of Python 3.9, use the exit_on_error parameter:

parser = argparse.ArgumentParser(exit_on_error=False)

This allows you to catch errors manually:

try:
    parser.parse_args('--integers a'.split())
except argparse.ArgumentError:
    print('Catching an argumentError')

answered Apr 13, 2021 at 10:35

Thomas

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Comments

larkost · Accepted Answer · 2018-01-08 21:30:18Z

-1

#!/usr/bin/env python
import argparse

parser = argparse.ArgumentParser()
url_command = parser.add_argument('url', metavar='URL', nargs='*', help='Specify URL(s)')
args = parser.parse_args()

url_command.nargs='+'  # format the usage correctly
if not args.url:
    parser.error('At least one URL is required!')

answered Jan 8, 2018 at 21:30

larkost

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1 Comment

Andrej Mohar Over a year ago

The question was how to suppress the error messages that argparse forces upon the user, not how to reformat the arguments so that the message is not thrown.

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